Calculate the carbon carbon double bond energy in ethane from the following reaction, H2C=CH2(g) + H2(g) –> H3C−CH3(g) ΔH = −138 kJ/mol. If Bond enthalpies are: C−C = 348; H−H = 436; C−H = 412 in KJ/mol.
Calculate the carbon carbon double bond energy in ethane from the following reaction, H2C=CH2(g) + H2(g) –> H3C−CH3(g) ΔH = −138 kJ/mol. If Bond enthalpies are: C−C = 348; H−H = 436; C−H = 412 in KJ/mol. Correct Answer 598 KJ/mol
By following the rule ΔHr = Σ ΔHf – ∑ ΔHf, we get -(x + 4(412) + 432) + (348 + 6(412)) = -138 KJ/mol; x = 598 KJ/mol. So the carbon carbon double bond energy in Ethane is given as 598 KJ per Mol.
মোঃ আরিফুল ইসলাম
Feb 20, 2025